Đặt:

\(X=\left(1+\dfrac{1}{9}\right)\left(1+\dfrac{1}{10}\right)\left(1+\dfrac{1}{11}\right).....\left(1+\dfrac{1}{200}\right)\)

\(X=\dfrac{10}{9}.\dfrac{11}{10}.\dfrac{12}{11}......\dfrac{201}{200}\)

\(X=\dfrac{10.11.12......201}{9.10.11......200}\)

\(X=\dfrac{201}{9}\)

\(Y=\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right).....\left(1-\dfrac{1}{99}\right)\)

\(Y=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}\)

\(Y=\dfrac{9.10.11......98}{10.11.12.....99}\)

\(Y=\dfrac{9}{99}=\dfrac{1}{11}\)

Giúp mình với mai mình phải nạp cho cô giáo

a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)

\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)

\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)

\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)

\(\Leftrightarrow x=\frac{-33}{28}\)

b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

a, .. . = 270

b, .. . = 135

c, 270 : 135 = 2

a) 1 . 5 . 6 + 2 . 10 . 12 = A 
     1 . 5 . 6 + 1 . 2 . 5 . 2 . 6 . 2 = A 
     1 . 5 . 6 . ( 1 + 2. 2 . 2 )       = A 
            30 . 9                           = A 
                      270   = A 
b) 1 . 3 . 5 + 2 . 6 . 10 = B 
      1 . 3 . 5 + 1 . 2 . 3 . 2 . 5 . 2 = B 
      1 . 3 . 5 . ( 1 + 2 . 2 . 2 )   = B 
            15 . 9  = B 
               135  = B 
c) A : B = A/B 

A, 11/12

B, 12,6

a,

x- 7/12 = 1/3

x = 1/3 + 7/12

x = 11/12

b,

x + 31/10 = 14,5

x = 14,5 - 31/10

x = 57/5

\(a,\\ x=\dfrac{1}{3}+\dfrac{7}{12}=\dfrac{11}{12}\\ b,\\ x+\dfrac{31}{10}=\dfrac{157}{10}\\ x=\dfrac{157}{10}-\dfrac{31}{10}=\dfrac{63}{5}\)

a) Quy đồng bỏ mẫu rồi giai pt ta đc : \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b)\(x=1\)

Con bai 2 thi sao a

\(\text{a, 3(x+1)+4x=10}\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow7x+3=10\)

\(\Rightarrow7x=10-3=7\)

\(\Rightarrow x=1\)

c, x+1/10+x+2/9=x+3/8+x+4/7

=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)

=> x+11/10 + x+11/9 = x+11/8 + x+11/7

...............

a) \(3\left(x+1\right)+4x=10\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow3x+4x=10-3\)

\(\Rightarrow7x=7\)

\(\Rightarrow x=7\)

\(A=\frac{9}{10}\times\frac{10}{11}\times\frac{11}{12}\times...\times\frac{99}{100}\)

\(A=\frac{9\times10\times11\times...99}{10\times11\times12\times...\times100}\)

\(\Rightarrow A=\frac{9}{100}\)

\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)

\(A=\left(-\frac{9}{10}\right)\cdot\left(-\frac{10}{11}\right)\cdot\left(-\frac{11}{12}\right)\cdot....\cdot\left(-\frac{98}{99}\right)\left(-\frac{99}{100}\right)\)

\(A=\frac{\left(-9\right)\left(-10\right)\left(-11\right)...\left(-98\right)\left(-99\right)}{10\cdot11\cdot12\cdot....\cdot99\cdot100}\)

\(A=\frac{9\cdot10\cdot11\cdot....\cdot98\cdot99}{10\cdot11\cdot12\cdot...\cdot99\cdot100}=\frac{9}{100}\)

Đáp án là \(-\frac{9}{100}\)nhá